Review question

# Can we show that this point moves on a circle? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6234

## Solution

1. The point $A$ has coordinates $(5,16)$ and the point $B$ has coordinates $(-4,4)$. The variable point $P$ has coordinates $(x,y)$ and moves on a path such that $AP = 2BP$. Show that the Cartesian equation of the path of $P$ is $(x + 7)^2 + y^2 = 100.$

We are told that $AP = 2BP$, which is the same as $(AP)^2 = 4(BP)^2$.

We use the formula for the distance $d$ between two points $(x_1,y_1)$ and $(x_2,y_2)$, which is $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2$.

So the above expression becomes \begin{align*} &&(x - 5)^2 + (y - 16)^2 &= 4((x + 4)^2 + (y - 4)^2)&&\quad\\ \iff\quad&& x^2 + y^2 - 10x - 32y + 281 &= 4x^2 + 4y^2 + 32x - 32y + 128\\ \iff\quad&& 3x^2 + 3y^2 + 42x - 153 &= 0\\ \iff\quad&& x^2 + y^2 + 14x - 51 &= 0. \end{align*} Completing the square for $x$ gives us \begin{align*} &&(x + 7)^2 - 49 + y^2 - 51 &= 0&&\quad\\ \iff\quad&& (x + 7)^2 + y^2 &= 100. \end{align*}

This is the equation of a circle with centre $(-7,0)$ and radius $10$.

1. The point $C$ has coordinates $(a,0)$ and the point $D$ has coordinates $(b,0)$. The variable point $Q$ moves on a path such that $QC = k \times QD,$ where $k > 1$. Given that the path of $Q$ is the same as the path of $P$, show that $\frac{a + 7}{b + 7} = \frac{a^2 + 51}{b^2 + 51}.$ Show further that $(a + 7)(b + 7) = 100$, in the case $a \neq b$.
Using the same approach as in part (i), with $Q=(x,y)$, we have \begin{align*} &&(QC)^2 &= k^2 \times (QD)^2&&\quad\\ \iff\quad&& (x - a)^2 + y^2 &= k^2(x - b)^2 + k^2y^2\\ \iff\quad&& x^2 - 2ax + a^2 + y^2 &= k^2 x^2 - 2bk^2 x + b^2 k^2 + k^2 y^2\\ \iff\quad&& x^2(k^2 - 1) + y^2(k^2 - 1) &+ x(2a - 2b k^2) + (b^2 k^2 - a^2) = 0. \end{align*}

If this locus is the same as the locus of $P$, which had equation $x^2+y^2+14x-51=0$, then we must be able to rewrite our equation in this form. (Alternatively, we can say that the ratios of corresponding coefficients must be the same.)

Note that $k^2 - 1 \neq 0$, so we can divide by it to make the coefficients of the $x^2$ and $y^2$ terms equal to $1$.

Dividing by $k^2-1$ gives the locus of $Q$ as $x^2 + y^2 + \frac{2a - 2b k^2}{k^2 - 1}x + \frac{b^2 k^2 - a^2}{k^2 - 1} = 0.$

If this locus is the same as the locus of $P$, then we must have $\frac{2a - 2b k^2}{k^2 - 1} = 14$ and $\frac{b^2 k^2 - a^2}{k^2 - 1} = -51.$

The first of these gives $2a - 2b k^2 = 14k^2 - 14,$ so $k^2 = \frac{a + 7}{b + 7}.$

Similarly, the second rearranges to

$b^2 k^2 - a^2 = -51k^2 + 51,$ so $k^2 = \frac{a^2 + 51}{b^2 + 51}.$

Thus $\frac{a + 7}{b + 7} = \frac{a^2 + 51}{b^2 + 51}.$

Now we can rearrange this to \begin{align*} &&(a + 7)(b^2 + 51) &= (b + 7)(a^2 + 51)&&\quad\\ \iff\quad&& ab^2 + 7b^2 + 51a &= a^2b + 7a^2 + 51b\\ \iff\quad&& ab^2 - a^2b &= 7(a^2 - b^2) + 51(b - a)\\ \iff\quad&& ab(b - a) &= 7(a - b)(a + b) + 51(b - a). \end{align*}

Since $a \neq b$, we may divide by $b-a$ to obtain $ab = 51 - 7(a + b).$

That is, \begin{align*} &&ab + 7(a + b) &= 51&&\quad\\ \Longrightarrow \quad&& ab + 7a + 7b + 49 &= 51 + 49\\ \Longrightarrow \quad&& (a + 7)(b + 7) &= 100. \end{align*}